Question

Let $a>0,b>0$. Let e and $\ell$ respectively be the eccentricity and length of the latus rectum of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Let $e^{\prime }$ and ${\ell }^{\prime }$ respectively the eccentricity and length of the latus rectum of its conjugate hyperbola. If $e^2=\frac{11}{14}\ell$ and ${\left(e^{\prime }\right)}^2=\frac{11}{8}{\ell }^{\prime }$, then the value of $77a+44b$ is equal to

• A. 100
• B. 110
• C. 120
• D. 130

Answer 1

Answer: (D)

$e=\sqrt{1+\frac{b^2}{a^2}},\ell =\frac{2b^2}{a}$
Given $e^2=\frac{11}{14}\ell$
$1+\frac{b^2}{a^2}=\frac{11}{14}\cdot \frac{2b^2}{a}$
$\frac{a^2+b^2}{a^2}=\frac{11}{7}\cdot \frac{b^2}{a}$.........(1)
Also $e^{\prime }=\sqrt{1+\frac{a^2}{b^2}},{\ell }^{\prime }=\frac{2a^2}{b}$
Given ${\left(e^{\prime }\right)}^2=\frac{11}{8}{\ell }^{\prime }$
$1+\frac{a^2}{b^2}=\frac{11}{8}\cdot \frac{2a^2}{b}$
$\frac{a^2+b^2}{b^2}=\frac{11}{4}\cdot \frac{a^2}{b}$
New (1) $÷(2)$
$\frac{b^2}{a^2}=\frac{4}{7}\cdot \frac{b^3}{a^3}$
$\therefore 7a=4b$
From (2)
$\frac{\frac{16b^2}{49}+b^2}{b^2}=\frac{11}{4}\cdot \frac{16b^2}{49b}$
$\frac{65}{49}=\frac{11}{4}\cdot \frac{16}{49}\cdot b$
$\therefore b=\frac{4\times 65}{11\times 16}$.......(4)
We have to find value of
$77a+44b$
$11(7a+4b)=11(4b+4b)=11\times 8b$
$\therefore$ Value of $11\times 8b=11\times 8\times \frac{4\times 65}{16\times 11}=130$